Question

How do I find the eigenvectors of the following matrix?

Answer 1

\[A=\left[\begin{matrix}1 & 2 \\ 2 & -2\end{matrix}\right]\]

Answer 2

\[Av = \lambda v\]

Answer 3

a=-6

Answer 4

first find eigenvalues :det(A - λI) = 0
λI - A = \[\left[\begin{matrix}1-λ & 2 \\ 2 & -2-λ\end{matrix}\right]\]

so Det :

(1−λ)(−2−λ) - 4 = 0
λ^2 - λ + 2λ -2 -4 = 0
λ^2 +λ -6 = 0
λ^2 +3λ - 2λ - 6 = 0
λ(λ+3) - 2(λ+3)
(λ-2)(λ+3) = 0

λ=-3,2

now

Av=λv

Av = 2v

v1+2v2 = 2v1
2v1 -2v2 = 2v2

any vector of the form : 2v2 = v1

Av = -3v

v1 + 2v2 = -3v1
2v1 - 2v2 = -3v2

2v2 = -4v1
any vector of the form : v2 = -2v1

Answer 5

by the way .. the eigenvector cant be zero.

Answer 6

example for the eigenvectors are :
2v2 = v1
-> (2,1)

v2 = -2v1
->(-1,2)

Answer 7

please tell me if it is fine..

Answer 8

Sorry - I have actually left my books for a moment... I will definitely check this when I get back to them - thanks so much!!

Answer 9

Totally forgot about this! Thank you so much - that was perfect!

Answer 10

oh im glad that you answered eventually :)

Answer 11

Haha, later is better than never, right? :P

Quick question, for the eigenvalue -3, is the final eigenvector\[v=\left(\begin{matrix}-2 \\ 1\end{matrix}\right)\]
Since \[v_2=-2v_1\]??

Answer 12

if you take v1 = 1 then v2 = -2

(1,-2)

its the opposite from yours

Answer 13

Oh I see, thanks!

Answer 14

So would the eigenspace of eigenvalue 2, be:
\[span \left\{ \left(\begin{matrix}2 \\ 0\end{matrix}\right) ,\left(\begin{matrix}0 \\ 1\end{matrix}\right)\right\}\]
or just \[span \left\{ \left(\begin{matrix}2 \\ 1\end{matrix}\right) \right\}\] Thanks for the ongoing help :)

Answer 15

Never mind, I got it :)